To solve this problem it is necessary to apply the concepts related to Period in a spring, elastic potential energy and energy in simple harmonic movement.
From the definition we know that the period can be expressed as
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
Where
m= Mass
k = Spring constant
Our values are given as
m = 16.6Kg
K = 2470N/m
Therefore the period would be
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi \sqrt{\frac{16.6}{2470}}[/tex]
[tex] T = 0.515s [/tex]
PART A) The frequency is the inverse of the period therefore
[tex]f = \frac{1}{T}\\f = \frac{1}{0.515}\\f = 1.9417Hz[/tex]
PART B) Elastic potential energy depends on compression and elasticity therefore
[tex]PE = \frac{1}{2}kx^2[/tex]
[tex]PE = \frac{1}{2} (2470)(0.281)^2[/tex]
[tex]PE = 97.51J[/tex]
PART C) Kinetic energy depends on mass and speed therefore
[tex]KE = \frac{1}{2}mv^2\\KE = \frac{1}{2} (16.6)(12.6)^2\\KE = 1371.7J[/tex]
PART D) As energy is conserved, the total energy is equivalent to the dual ratio of the elasticity constant and the amplitude, mathematically,
[tex]KE+PE = \frac{1}{2}kA^2[/tex]
[tex]1371.7+97.51 = \frac{1}{2}2470A^2[/tex]
[tex]A = \sqrt{\frac{1415*2}{2470}}[/tex]
[tex]A = 1.07m[/tex]
Therefore the motion's amplitude is 1.07m
